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Finding value of $\displaystyle \lim_{n\rightarrow \infty}\bigg(\frac{(kn)!}{n^{kn}}\bigg)^{\frac{1}{n}}$ for all $k>1$

Try: I have solved it using stirling Approximation

$\displaystyle n!\approx \bigg(\frac{n}{e}\bigg)^n\sqrt{2\pi n}$ for laege $n$

So we have $\displaystyle \lim_{n\rightarrow \infty}\bigg(\frac{kn}{e}\bigg)^{kn}\cdot \bigg(\sqrt{2\pi k n}\bigg)^{\frac{1}{n}}\cdot \frac{1}{n^k}=\bigg(\frac{k}{e}\bigg)^k$

Could some help me how to solve it without

stirling Approximation

Thanks.

Mr Pie
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DXT
  • 12,047

2 Answers2

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Simply note that this is a Riemann sum for positive $k$.

Letting $$\lim_{n\to\infty}\left(\frac{(kn!)}{n^{kn}}\right)^{1/n}=y\,,$$ we have

\begin{align} \ln y&=\lim_{n\to\infty}\frac{1}{n}\ln\left(\frac{(kn)!}{n^{kn}}\right) \\&=\lim_{n\to\infty}\frac{1}{n}\ln\left(\prod_{j=1}^{kn}\frac{j}n\right) \\&=\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{kn}\ln\left(\frac{j}n\right) \\&=\int_0^{k}\ln x\,dx=\ln\left(\frac{k}e\right)^k \end{align}

We can interchange log and limit as $\ln(\cdot)$ is continuous in its domain.

StubbornAtom
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$$ \lim_{n\rightarrow \infty} \left(\dfrac{(kn)!}{n^{kn}} \right)^{\dfrac{1}{n}} = \lim_{n\rightarrow \infty} \left(\dfrac{(kn)!}{\dfrac {kn^{kn}}{k}} \right)^{\dfrac{k}{kn}} = {(\lim_{n\rightarrow \infty} \left(\dfrac{(kn)!}{\dfrac {kn^{kn}}{k}} \right)^{\dfrac{1}{kn}}})^{k} = \left(\dfrac{k}{e} \right)^k $$

The last step is derived from: $$y =\lim_{n\rightarrow \infty}\bigg(\dfrac{(n)!}{n^{n}}\bigg)^{\dfrac{1}{n}}$$ $\ln{y} =\displaystyle \lim_{n\rightarrow \infty} 1/n × \ln(1/n) × \ln(2/n) × ... × \ln(1)$

= $\sum_{r=1}^{n}1/n × \ln{r/n}$

= $\int_0^{1}\ln x\,dx$ = $-1$

(Converting infinite sum into an integral) And therefore, $y = e^{-1}$. Plug in the same above.

Iceberry
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